Derivative of the Lorentz factor

  • #1
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According to David Morin (link: https://books.google.com/books?id=Ni6CD7K2X4MC&pg=PA636), the time-derivative of the Lorentz factor is (##c=1##):

##\dot{\gamma} = \gamma^3 v \dot{v}##,

and the four-acceleration:

##\mathbf{A} = (\gamma^4 v \dot{v}, \gamma^4 v \dot{v} \mathbf{v} + \gamma^2 \mathbf{a})##.

Is that a mistake? I thought that it's:

##\dot{\gamma} = \gamma^3 (\mathbf{v} \cdot \dot{\mathbf{v}})##,

and

##\mathbf{A} = (\gamma^4 (\mathbf{v} \cdot \dot{\mathbf{v}}), \, \gamma^4 (\mathbf{v} \cdot \dot{\mathbf{v}}) \mathbf{v} + \gamma^2 \mathbf{a})##.


I could be wrong, but my understanding is that the ##v^2## in the radicand is really ##\mathbf{v} \cdot \mathbf{v}##, so that:

##\dfrac{d}{dt} \, (1 - v^2)^{-1/2} = \dfrac{d}{d(v^2)} \, (1 - v^2)^{-1/2} \, \dfrac{d}{dt} \, (\mathbf{v} \cdot \mathbf{v})##.
 

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  • #2
Orodruin
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Since ##v^2 = \vec v \cdot \vec v##, it follows directly that ##v \dot v = \vec v \cdot \dot{\vec v}##. However, note that ##\dot v## is generally not equal to ##|\dot{\vec v}|##.
 
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